Improper Integrals
- Page ID
- 218547
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Improper Integrals Integrals with Infinite Limits Up until now, all the integrals that we have performed have had finite limits and integrands with no asymptotes. Now we will investigate integrals with infinite limits and integrands with vertical asymptotes. We begin with infinite limits.
Note: A similar definition applies to integral with negative infinity as a limit.
Example Evaluate \( \int_1^{\infty} \frac{1}{x^2}dx \)
Solution We use the definition to find \( \lim\limits_{M \to \infty} \int_1^M \frac{1}{x^2}dx = \lim\limits_{M \to \infty}(\frac{-1}{x})_1^M = \lim\limits_{M \to \infty} (\frac{-1}{M} + 1) = 1 \) Integrals with Vertical Asymptotes The fundamental theorem of calculus tells us how to evaluate integrals if the functions is continuous. Id does not apply if the function has a vertical asymptote. In this case we will need to make an adjustment. The definition below tells us how the make this adjustment.
Example The integrand has a vertical asymptote at x = 0, so we must use part B if the definition. We have
Application Statisticians are always taking about the normal or bell curve. This curve is defined by the function \( f(x) = \frac{1}{\sigma \sqrt{2\pi}}e^{\frac{-(x - \mu)^2}{2 \sigma^2}} \) Where m is the mean of the distribution and s is the standard deviation. With a probability distribution of f(x), the probability of an event occurring between a and b is given by \( P(a \le x \le b) = \int_a^b f(x) dx \)
Example The weight of a species of rodent is normally distributed with mean 0.5 kg and standard deviation 0.1 kg. What is the probability of a randomly selected specimen to have a weight of greater then 0.6 kg?
Solution Since we are not given an upper bound on the weight, we use infinity for the upper limit. We have \( P(0.6 \le x \le \infty) = \int_{0.6}^{\infty} \frac{1}{0.1 \sqrt{2\pi}}e^{\frac{-(x - 0.5)^2}{2 0.1^2}}dx = 3.99 \int_{0.6}^{\infty} e^{-50(x - 0.5)^2} dx \) This is an improper integral such that there is no simple antiderivative. We use an approximation technique to solve it such as the Midpoint Rule or Simpson's Rule or a calculator to get an answer of 0.1586. Author's Note: I put this into the TI 89 calculator which produced the above number with a warning of "questionable accuracy".
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